Question: $h(r)=-(r+9)^2+36$ 1) What are the zeros of the function? Write the smaller $r$ first, and the larger $r$ second. $\text{smaller }r=$
$\begin{aligned} -(r+9)^2+36&=0 \\\\ -(r+9)^2&=-36 \\\\ (r+9)^2&=36 \\\\ \sqrt{(r+9)^2}&=\sqrt{36} \\\\ r+9&=\pm 6 \\\\ r&=\pm6-9 \\\\ r={-15}&\text{ or }r={-3} \end{aligned}$ $h(r)$ is given in vertex form: $h(r)=-(r-({-9}))^2+{36}$ So the vertex of the parabola is at $({-9},{36})$. In conclusion, $\begin{aligned} \text{smaller }r&=-15 \\\\ \text{larger }r&=-3 \end{aligned}$ The vertex of the parabola is at $(-9,36)$